Solutions Test -22

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Solutions Test -22

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Question 1
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In isotonic solutions .......

(i) solute and solvent both are same.

(ii) osmotic pressure is same.

(iii) solute and solvent may or may not be same.

(iv) solute is always same solvent may be different.

A
(ii) and (iii)

B
(iii) and (i)

C
(i) and (iv)

D
(ii) and (i)

Solution

The two solutions having same osmotic pressure are known as isotonic solutions. The solute and solvent particles may or may not be same but osmotic pressure must be same.
Question 2
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Colligative properties are observed when

(i) a nonvolatile solid is dissolved in a volatile liquid.

(ii) a nonvolatile liquid is dissolved in another volatile liquid.

(iii) a gas is dissolved in nonvolatile liquid.

(iv) a volatile liquid is dissolved in another volatile liquid.

A
(i) and (ii)

B
(ii) and (iii)

C
(iv) and (ii)

D
(i) and (iii)

Solution

When any of one component of binary mixture either solvent or solute is volatile it causes deviation from ideal behaviour and vapour pressure of solution which causes change in colligative property.

Question 3

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Match the items given in Column I with the type of solutions given in Column II.

Column I	Column II
(i) Soda water	(a) A solution of gas in solid
(ii) Sugar solution	(b) A solution of gas in gas
(iii) German silver	(c) A solution of solid in liquid
(iv) Air	(d) A solution of solid in solid
(v) Hydrogen gas in palladium	(e) A solution of gas in liquid
	(f) A solution of liquid in solid

(i)- (e) (ii)- (c) (iii)- (d) (iv)- (b) (v)- (a)

(i)- (a) (ii)- (b) (iii)- (c) (iv)- (d) (v)- (e)

(i)- (e) (ii)- (a) (iii)- (b) (iv)- (c) (v)- (d)

(i)- (a) (ii)- (b) (iii)- (c) (iv)- (d) (v)- (e)

Solution

A. Soda water A solution of gas in liquid, e.g., $CO_2$ in soft drinks.

B. Sugar solution A solution of solid in liquid in which sugar particles (solid) are dissolved in water (liquid).

C. German silver German silver is an alloy which is a solid solution of solid in solid. It is an alloy of Cu, Zn and Ni.

D. Air A solution of gas in gas. Air is a mixture of various gases.

E. Hydrogen gas in palladium is an example of solution of gas in solid. This is used as an reducing agent

Question 4
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Assertion : When NaCl is added to water a depression in freezing point is observed.

Reason : The lowering of vapour pressure of a solution causes depression in the freezing point.

A
Assertion and reason both are correct statements and reason is correct explanation for assertion.

B
Assertion and reason both are correct statements but reason is not correct explanation for assertion.

C
Assertion is correct statement but reason is wrong statement.

D
Assertion and reason both are incorrect statements.

Solution

On addition of nonvolatile solute (viz. NaCl) to water NaCl solution is formed.Due to relatively lesser number of water molecules at the surface of liquid , the solution exerts a lower vapour pressure as compared to that of pure water. It is because of this lowering of vapour pressure that a depression in freezing point of water is observed.
Question 5
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Assertion : When a solution is separated from the pure solvent by a semipermeable membrane, the solvent molecules pass through it from pure solvent side to the solution side.

Reason : Diffusion of solvent occurs from a region of high concentration solution to a region of low concentration solution.

A
Assertion and reason both are correct statements and reason is correct explanation for assertion.

B
Assertion and reason both are correct statements but reason is not correct explanation for assertion.

C
Assertion is correct statement but reason is wrong statement.

D
Assertion and reason both are incorrect statements.

Solution

Assertion is correct statement but reason is a wrong statement because a semipermeable membrane permits solvent molecules to pass through a solution of lower concentration to that of higher concentration Flow of solvent molecule from solvent side to solution side through semipermeable membrane is called osmosis.
Question 6
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On the basis of information given below mark the correct option.

Information :

(A) In bromoethane and chloroethane mixture intermolecular interactions of A–A and B–B type are nearly same as A–B type interactions.

(B) In ethanol and acetone mixture A–A or B–B type intermolecular interactions are stronger than A–B type interactions.

(C) In chloroform and acetone mixture A–A or B–B type intermolecular interactions are weaker than A–B type interactions.

A
Solution (B) and (C) will follow Raoult’s law.

B
Solution (A) will follow Raoult’s law.

C
Solution (B) will show negative deviation from Raoult’s law.

D
Solution (C) will show positive deviation from Raoult’s law.

Solution

For an ideal solution A-A and B-B intermolecular interactions should be nearly same as A-B type interactions.
Question 7
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If two liquids A and B form minimum boiling azeotrope at some specific composition then ..................

A
A–B interactions are stronger than those between A–A or B–B.

B
vapour pressure of solution increases because more number of molecules of liquids A and B can escape from the solution.

C
vapour pressure of solution decreases because less number of molecules of only one of the liquids escape from the solution.

D
A–B interactions are weaker than those between A–A or B–B

Solution

The solutions which show large positive deviation from Rault's law form minimum boiling azeotrope at some specific composition . (ii) In case of positive deviation from Rault's law A-B interactions are weaker than those between A-A or B-B.
Question 8
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4L of 0.02 M aqueous solution of NaCl was diluted by adding one litre of water. The molality of the resultant solution is

A
0.004

B
0.008

C
0.012

D
0.016

Solution

Apply the relation : $M_1\,V_1 \,= \,M_2\,\,V_2 $

Given:

$M_1\,=\,0.02\,M,\,V_1\,=\,4L,\,M_2\,=\,? \,V_2\,=\,5L$

$therefore, 0.02\,\times 4L = M_2\,\times\, 5L$

$ M_2\,=\,0.08/5$

= 0.016 M
Question 9
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On the basis of information given below mark the correct option.

Information : On adding acetone to methanol some of the hydrogen bonds between methanol molecules break.

A
At specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show positive deviation from Raoult’s law.

B
At specific composition methanol-acetone mixture forms maximum boiling azeotrope and will show positive deviation from Raoult’s law.

C
At specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show negative deviation from Raoult’s law.

D
At specific composition methanol-acetone mixture will form maximum boiling azeotrope and will show negative deviation from Raoult’s law.

Solution

(i) ( A-A) or( B-B ) interactions are stronger than the( A-B ) interactions ; where, A is methanol molecule & B represents a molecule of acetone.It means that in this solution molecules of A (or B ) will find it easier to escape. This will increase the vapour pressure and result into positive deviation from Rault's law .Further,

(ii) due to this positive deviation the methanol - acetone mixture forms minimum boiling azeotrope.
Question 10
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Choose the correct statement with respect to the vapour pressure of a liquid among the following.

A
Increases linearly with increasing temperature.

B
Increases non-linearly with increasing temperature.

C
Decreases linearly with increasing temperature.

D
Decreases non-linearly with increasing temperature.

Solution

Vapour pressure is defined as the pressure exerted by the vapours above the liquid surface in equilibrium with the liquid at a given temperature. The vapour pressure of a liquid increases non-linearly with increasing temperature.

This is because kinetic energy is the function of temperature which means that as the temperature is increased, more molecules will have greater kinetic energies and thus they can escape from the surface of the liquid to the vapour phase resulting in higher vapour pressure.

Option B is correct.
Question 11
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Two $$5$$ molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are $$M_X$$ and $$M_Y$$, respectively where $$M_X=\displaystyle\frac{3}{4}M_Y$$. The relative lowering of vapour pressure of the solution in X is "m" times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of solvent, the value of "m" is?

A
$$\displaystyle\frac{3}{4}$$

B
$$\displaystyle\frac{1}{2}$$

C
$$\displaystyle\frac{1}{4}$$

D
$$\displaystyle\frac{4}{3}$$

Solution

THe relationship between molar masses of two solvents is
$$\displaystyle M_X=\dfrac {3}{4}M_Y$$......(1)
THe relative lowering of vapour pressure of two solutions is
$$\displaystyle (\dfrac{\Delta P}{P})_X=m(\dfrac{\Delta P}{P})_Y$$
But, the relative lowering of vapour pressure of solution is directly proportional to the mole fraction of solute.
$$\displaystyle M_x \times \dfrac {5}{1000} = m \times M_Y \times \dfrac {5}{1000}$$.....(2)
Substitute equation (1) in equation (2).
$$\displaystyle \dfrac {3}{4} \times M_Y \times \dfrac {5}{1000} = m \times M_Y \times \dfrac {5}{1000}$$
$$\displaystyle m=\dfrac {3}{4}$$
Note:
5 molal solution means 5 moles of solute are dissolved in 1 kg (or 1000 g) of solvent. The number of moles of solvent $$\displaystyle = \dfrac {1000g}{M} $$
The mole fraction of solute
$$\displaystyle = \dfrac {5}{1000/M}$$
$$\displaystyle = M \times \dfrac {5}{1000}$$
Question 12
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For an ideal solution of two components A and B, which of the following is true?

A
$$\Delta H_{mixing} < 0(zero)$$

B
$$\Delta H_{mixing} > 0(zero)$$

C
A - B interaction is stronger than A - A and B - B interactions

D
A - A, B - B and A - B interactions are identical

Solution

The solutions which obey Raoult's law over the entire range of concentration are known as ideal solutions. The ideal solutions have two other important properties. The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero.

At a molecular level, ideal behaviour of the solutions can be explained by considering two components A and B. In pure components, the intermolecular attractive interactions will be of types A-A and B-B, whereas in the binary solutions in addition to these two interactions, A-B type of interactions will also be present. If the intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B, this leads to the formation of the ideal solution.

Hence, as the given solution is ideal, the correct option is $$(D)$$.
Question 13
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Liquid $$M$$ and liquid $$N$$ form an ideal solution. The vapour pressures of pure liquids $$M$$ and $$N$$ are $$450$$ and $$700\ mm$$ $$Hg$$, respectively, at the same temperature. Then correct statement is:
($${x}_{M}=$$ Mole fraction of $$M$$ in solution;
$${x}_{N}=$$ Mole fraction of $$N$$ in solution;
$${y}_{M}=$$ Mole fraction of $$M$$ in vapour phase;
$${y}_{N}=$$ Mole fraction of $$N$$ in vapour phase)

A
$$({x}_{M}-{y}_{M})< ({x}_{N}-{y}_{N})$$

B
$$\cfrac{{x}_{M}}{{x}_{N}}< \cfrac{{y}_{M}}{{y}_{N}}$$

C
$$\cfrac{{x}_{M}}{{x}_{N}}> \cfrac{{y}_{M}}{{y}_{N}}$$

D
$$\cfrac{{x}_{M}}{{x}_{N}}= \cfrac{{y}_{M}}{{y}_{N}}$$

Solution

Since $$ P^o_N>P^o_M,$$ so $$N$$ is more volatile than $$M.$$

$$\therefore y_N>x_N$$

and $$x_M>y_M$$

Multiply we get

$$y_N\times x_M>x_N \times y_M$$

$$i.e.\ \dfrac{x_M}{x_N}=\dfrac{y_M}{y_N}$$
Question 14
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The observed osmotic pressure for a 0.10 M solution of $$Fe(NH_4)_2(SO_4)_2$$ at $$25^oC$$ is 10.8 atm. The expected and experimental (observed) values of Van't Hoff factor (i) will be respectively:

$$[R=0.082\ L\ {atm} K^{-1} mol^{-1}]$$

A
$$5$$ and $$4.42$$

B
$$4$$ and $$4.00$$

C
$$5$$ and $$3.42$$

D
$$3$$ and $$5.42$$

Solution

$$Fe(NH_4)_2(SO_4)_2 \longrightarrow 2NH_4^+ + Fe^{2+} + 2{SO_4}^{2-}$$

Expected value of $$i = 5$$

Now, observed van't hoff factor $$= \displaystyle \dfrac{P}{MRT}$$

$$= \dfrac{10.8}{0.10 \times 0.082 \times 298}$$

$$= 4.42$$

Hence, option A is correct.
Question 15
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Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non volatile solute are together sealed in a container. Over time:

A
The volume of the solution does not change and the volume of the solvent decreases

B
the volume of the solution deceases and the volume of the solvent increases

C
the volume of the solution and the solvent does not change

D
the volume of the solution increases and the volume of the solvent decreases

Solution

As solvent is pure and volatile, so it will vapourise and increase vapour pressure till that vapour pressure will come in equilibrium so volume decrease.

As in solution, there is non-volatile solute, so it will not vapourise for vapour pressure to be in equilibrium solvent vapour will come towards the solution. So, the volume of solution increase.

So, the volume of the solution increases and the volume of the solvent decreases.
Question 16
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Consider the following liquid - vapour equilibrium Liquid $$\rightleftharpoons$$ Vapour. Which of the following relations is correct?

A
$$\dfrac {dlnG}{dT^{2}} = \dfrac {\Delta H_{v}}{RT^{2}}$$

B
$$\dfrac {dlnP}{dT} = \dfrac {-\Delta H_{v}}{RT}$$

C
$$\dfrac {dlnP}{dT^{2}} = \dfrac {-\Delta H_{v}}{T^{2}}$$

D
$$\dfrac {dlnP}{dT} = \dfrac {\Delta H_{v}}{RT^{2}}$$

Solution

According to Clausius - Clapeyron's equation
$$\dfrac {d \ lnP}{dT}= \dfrac {\Delta H_v}{RT^2}$$
According to this equation, the rate at which the natural logarithm of the vapor pressure of a liquid changes with temperature is determined by the molar enthalpy of vaporization of the liquid, the ideal gas constant, and the temperature of the system.
Question 17
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Which one of the following electrolytes has the same value of van't Hoff's factor (i) as that of $$Al_2(SO_4)_3$$ (if all are 100% ionised)?

A
$$Al(NO_3)_3$$

B
$$K_4[Fe(CN)_6]$$

C
$$K_2SO_4$$

D
$$K_3[Fe(CN)_6]$$

Solution

Van't Hoff factor $$i =\dfrac {\text{number of solute particles present in solution}}{\text{theoretical number of solute particles due to solution of non electrolyte}} =\dfrac {n(observed)}{n(theoretical)}$$.

1 molecule of $$Al_2(SO_4)_3$$ ionizes in solution to produce 5 ions.

$$Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$$

Hence, $$i=\dfrac {n(observed)}{n(theoretical)}=\dfrac {5}{1}=5$$.

1 molecule of $$K_4[Fe(CN)_6]$$ ionize in solution to produce 5 ions.

$$K_4[Fe(CN)_6] \rightarrow 4K^+ +[Fe(CN)_6]^{4-}$$

Hence, $$i=\dfrac {n(observed)}{n(theoretical)}=\dfrac {5}{1}=5$$.

Thus, $$K_4[Fe(CN)_6]$$ has the same value of van't Hoff's factor (i) as that of $$Al_2(SO_4)_3$$ (assuming 100% ionization).

Note: 1 molecule each of $$Al(NO_3)_3$$, $$K_2SO_4$$ and $$K_3[Fe(CN)_6]$$ ionizes in solution to produce 4, 3 and 4 ions respectively.
Question 18
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For an ideal solution, the correct option is :

A
$$\Delta_{mix}S=0$$ at constant T and P

B
$$\Delta_{mix}V\neq 0$$ at correct T and P

C
$$\Delta_{mix}H=0$$ at constant T and P

D
$$\Delta_{mix} G=0$$ at constant T and P

Solution

The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions.

The ideal solutions have two other important properties. The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero, i.e.,
For ideal solution,
$$\Delta_{mix}H=0$$ $$\Delta_{mix}V=0$$
Question 19
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Which one is not equal to zero for an ideal solution?

A
$$\Delta V_{mix}$$

B
$$\Delta P=P_{observed}-P_{Raoult}$$

C
$$\Delta H_{mix}$$

D
$$\Delta S_{mix}$$

Solution

For an ideal solution,

(A) The volume of mixing is zero. $$\Delta V_{mix}=0$$

(B) The observed pressure is equal to the pressure calculated from Raoult's law, then solution said to be ideal solution.

$$P_{observed}=P_{Raoult}$$

$$\Delta P=P_{observed}-P_{Raoult}=0$$

(C) The enthalpy change of mixing is zero $$\Delta H_{mix}=0$$.

(D) But the entropy of mixing is not equal to zero. $$\Delta S_{mix} \neq 0$$.

Option D is correct.

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Solutions Test -22

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Solutions Test -22

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Question 1

(ii) and (iii)

(iii) and (i)

(i) and (iv)

(ii) and (i)

Solution

Question 2

(i) and (ii)

(ii) and (iii)

(iv) and (ii)

(i) and (iii)

Solution

Question 3

(i)- (e) (ii)- (c) (iii)- (d) (iv)- (b) (v)- (a)

(i)- (a) (ii)- (b) (iii)- (c) (iv)- (d) (v)- (e)

(i)- (e) (ii)- (a) (iii)- (b) (iv)- (c) (v)- (d)

(i)- (a) (ii)- (b) (iii)- (c) (iv)- (d) (v)- (e)

Solution

Question 4

Assertion and reason both are correct statements and reason is correct explanation for assertion.

Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Assertion is correct statement but reason is wrong statement.

Assertion and reason both are incorrect statements.

Solution

Question 5

Assertion and reason both are correct statements and reason is correct explanation for assertion.

Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Assertion is correct statement but reason is wrong statement.

Assertion and reason both are incorrect statements.

Solution

Question 6

Solution (B) and (C) will follow Raoult’s law.

Solution (A) will follow Raoult’s law.

Solution (B) will show negative deviation from Raoult’s law.

Solution (C) will show positive deviation from Raoult’s law.

Solution

Question 7

A–B interactions are stronger than those between A–A or B–B.

vapour pressure of solution increases because more number of molecules of liquids A and B can escape from the solution.

vapour pressure of solution decreases because less number of molecules of only one of the liquids escape from the solution.

A–B interactions are weaker than those between A–A or B–B

Solution

Question 8

0.004

0.008

0.012

0.016

Solution

Question 9

At specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show positive deviation from Raoult’s law.

At specific composition methanol-acetone mixture forms maximum boiling azeotrope and will show positive deviation from Raoult’s law.

At specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show negative deviation from Raoult’s law.

At specific composition methanol-acetone mixture will form maximum boiling azeotrope and will show negative deviation from Raoult’s law.

Solution

Question 10

Increases linearly with increasing temperature.

Increases non-linearly with increasing temperature.

Decreases linearly with increasing temperature.

Decreases non-linearly with increasing temperature.

Solution

Question 11

$$\displaystyle\frac{3}{4}$$

$$\displaystyle\frac{1}{2}$$

$$\displaystyle\frac{1}{4}$$

$$\displaystyle\frac{4}{3}$$

Solution

Question 12

$$\Delta H_{mixing} < 0(zero)$$

$$\Delta H_{mixing} > 0(zero)$$

A - B interaction is stronger than A - A and B - B interactions

A - A, B - B and A - B interactions are identical

Solution